# First assignment using Integration by Parts

I'm doing my first assignment where I think we're required to use IBP... but i'm not 100% certain. u-substitution doesn't seem to work for this problem.

$$\int x^3e^{-2x}\ dx$$

I haven't gone all the way, but I do think IBP will work if I do it three times, but I hardly think that this was the professor's intention. Or maybe I am doing something wrong..

$$\int u\ dv=uv-\int v\ du$$

So given the question above, I've set

\begin{array}{cc} u=x^3 & v=-\frac{e^{-2x}}{2} \\ du=3x^2 & dv=e^{-2x}\ dx \end{array}

When plugging in these values now, I get $$\int x^3\cdot e^{-2x}\ dx=-\frac{x^3e^{-2x}}{2}+\int \frac{e^{-2x}\cdot 3x^2}{2}\ dx$$

And this doesn't seem to be any easier than before.

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Well if it doesn't work with one time what about trying it 2 or three times ? – Dominic Michaelis Feb 27 '13 at 18:04
My point is that it might work, but I don't think it was intended that I have to do it more than once or twice... It's the very first question on the assignment. I'm wondering if I did some calculation wrong, or if there's another way to do it before I assume that I have to do it three times. – agent154 Feb 27 '13 at 18:06
You know you're making progress because your algebraic term is on the order of $2$ now, namely $3x^2$, rather than order $3$ (the $x^3$). So, apply IBP a few more times - the power should go down by $1$ each time. Edit: after seeing your comment and a quick glance (about to head out the door), looks like IBP is required more than once from the natural choice of $u = x^3$. – Joe Feb 27 '13 at 18:06

I think you will have to do it multiple times. Here's the intuition: each time you integrate by parts, you can reduce the degree of the $x^n$ term by $1$. If you do it enough, eventually you are left with just the integral of $e^x$ (times some constant), which you know how to integrate. Then putting everything together gives you the answer. I don't think there is an easier or more natural way.